$\begin{align}  & 3\sin 3\theta -\operatorname{cosec}3\theta +2=0 \\ & 3\sin 3\theta -\frac{1}{\sin 3\theta }+2=0 \\ & 3{{\sin }^{2}}3\theta +2\sin 3\theta -1=0 \\ & 3{{\sin }^{2}}3\theta +3\sin 3\theta -\sin 3\theta -1=0 \\ & 3\sin 3\theta (\sin 3\theta +1)-1(\sin 3\theta +1)=0 \\ & (3\sin 3\theta -1)(\sin 3\theta +1)=0 \\ & \sin 3\theta =-1\text{ or }\sin 3\theta =\frac{1}{3} \\ & 3\theta ={{\sin }^{-1}}1=-{{90}^{\circ }} \\ & \text{The general solution is} \\ & 3\theta =n180+{{(-1)}^{n}}(-{{90}^{\circ }}) \\ & \theta =n{{60}^{\circ }}+{{(-1)}^{n}}(-{{30}^{\circ }}) \\ & \theta ={{270}^{\circ }} \\ & \text{Also,}\sin 3\theta =\frac{1}{3},\text{  }3\theta ={{\sin }^{-1}}\frac{1}{3}={{19}^{\circ }}.47' \\ & \text{The general solution will be } \\ & 3\theta =n180+{{(-1)}^{n}}({{19}^{\circ }}.47') \\ & \theta =\frac{n180+{{(-1)}^{n}}({{19}^{\circ }}.47')}{3} \\ & \theta ={{6.49}^{\circ }},{{53.51}^{\circ }},{{126.49}^{\circ }} \\ & \text{The value of }\theta \text{ are }{{6.49}^{\circ }},{{53.51}^{\circ }},126.49{{,}^{\circ }}{{270}^{\circ }} \\\end{align}$